Description

Problem C

Input: standard input

Output: standard output

Given two polynomials f(x) and g(x) in
Z_n, you have to find their GCD polynomial, ie, a polynomial
r(x) (also in Z_n) which has the greatest degree of all the polynomials in
Z_n that divide both f(x) and
g(x). There can be more than one such polynomial, of which you are to find the one with a leading coefficient of
1 (1 is the unity in Z_n. Such polynomial is also called a
monic polynomial).

(Note: A function f(x) is in Z_n means all the coefficients in
f(x) is modulo n.)

Input

There will be no more than 101 test cases. Each test case consists of three lines: the first line has
n, which will be a prime number not more than 1500. The second and third lines give the two polynomials
f(x) and g(x). The polynomials are represented by first an integer
D which represents the degree of the polynomial, followed by
(D + 1) positive integers representing the coefficients of the polynomial. the coefficients are in decreasing order of Exponent. Input ends with
n = 0. The value of D won‘t be more than
100.

Output

For each test case, print the test case number and r(x), in the same format as the input

Sample Input                                 Output for Sample Input

3 

3 2 2 1 1

4 1 0 2 22

0 

Case 1: 2 1 2 1 

Problem setter: SadrulHabibChowdhury

Special Thanks: Derek Kisman, EPS

Note: The first sample input has 2x³ + 2x² + x + 1
and x⁴ + 2x² + 2x + 2 as the functions.

题意：求两个多项式的最大公共多项式

思路：套用了模板

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
typedef long long ll;
using namespace std;
const int maxn = 100005;

vector<int> G[maxn];
int mod;

int pow_mod(int a, int b) {
	int ans = 1;
	while (b) {
		if (b & 1)
			ans = ans * a % mod;
		b >>= 1;
		a = a * a % mod;
	}
	return ans;
}

/*多项式求最大公共项*/
vector<int> poly_gcd(vector<int> a,vector<int> b) {
	if (b.size() == 0)
		return a;
	int t = a.size() - b.size();
	vector<int> c;
	for (int i = 0;i <= t; i++) {
		int tmp =  a[i] * pow_mod(b[0],mod-2)%mod;
		for (ll j = 0; j < b.size(); j++)
			a[i+j] = (a[i+j] - tmp * b[j]%mod + mod)%mod;
	}
	int p = -1;
	for (int i = 0;i < a.size(); i++) {
		if (a[i] != 0) {
			p = i;
			break;
		}
	}
	if (p >= 0) {
		for (int i = p; i < a.size(); i++)
			c.push_back(a[i]);
	}
	return poly_gcd(b,c);
}  

int main() {
	int cas = 1;
	while (scanf("%d", &mod) != EOF && mod) {
		for (int i = 0; i < 2; i++)
			G[i].clear();
		int a, b;

		for (int i = 0; i < 2; i++) {
			scanf("%d", &a);
			for (int j = 0; j <= a; j++) {
				scanf("%d", &b);
				G[i].push_back(b);
			}
		}

		vector<int> ans = poly_gcd(G[0], G[1]);
		printf("Case %d: %d", cas++, ans.size()-1);
		int cnt = pow_mod(ans[0], mod-2);
		for (int i = 0; i < ans.size(); i++) {
			ans[i] = ans[i] * cnt % mod;
			printf(" %d", ans[i]);
		}
		printf("\n");
	}
	return 0;
}